@Krijn: it's a she. I don't know, but she's not officially a student yet. the first semester is basically for getting used to the school and language, and the second semester is a probation time. if all goes well, she'll be an official student of the school in a year.
@user1618033 reminds of somone we were competing who gives the more accurately big probability ever existed or not he gives a thing like 99,9999999% i gave the 99.... root of the same proportion he can reach ever he didnt understand so i told him that i gave a thing only mathematicians do understand
Just for the record, before getting very involved in the area of the calculations of integrals, series and limits, I was doing all kind of Olympiad problems day by day, also including such problems (this particular one is a problem given in some Olympiad).
I think it helped me very much, and I entered pretty strong (in the sense of developing good analytical skills) in the area I'm presently interested in.
I'm highly fascinated by clever solutions in any area of mathematics (of course, I don't count the ones I didn't study).
Consider $\DeclareMathOperator{\Z}{\Bbb Z}\Spec \Z[i]\to\Spec\Z$ given by the inclusion of $\Z$ into $\Z[i]$.
now the fiber over every prime which is 1 mod 4 has two points, e.g. $f^{-1}((5)) = \{(1+2i),(1-2i)\}$ over primes congruent to 3 mod 4 (or over 2), there is one point.
in this pdf (not his main notes), Vakil asks the reader to compare the degree of the "residue" field extension with the number of points in the fiber. I'm guessing these will be the same except at $2$, but:
the weird thing is, isn't $\Z[i]/(3)$ isomorphic to $\Bbb F_{3^2}$?, since $x^2+1$ is irred mod $3$
@Krijn, if you're studying class field theory in the near future, I'd love to join you :)
The geometric picture here is unclear. In topology, I'd define ramified points to be the ones which have preimage cardinality less than a generic preimage cardinality (in this case, 2). So I would call 3 ramified. What's the definition here, again?
I forgot. (I think I have run into this issue before)
because if $X$ is compact then $X = \bigcup_{i=1}^{m} U_{i}$, then is clearly bounded and if we take the complement of X, that is, $\varline{X}$ we got $\emptyset$ which is strictily open in a bounded space, then $X$ is closed. Indeed, $X$ si compact. Right?
I proved the sequential definition of continuity is equivalent to epsilon-delta, I proved closed subsets of compact spaces are compact, and that O(n) is comact
